Question

If $|x|<1$, then $\underset{n\to \infty }{lim}\left\{\right(1+x\left)\right(1+x^2\left)\right(1+x^4).....(1+x^{2n}\left)\right\}$ is equal to

• A. $\frac{1}{x-1}$
• B. $\frac{1}{1-x}$
• C. $1-x$
• D. $x-1$

Answer 1

Answer: (B)

$\frac{1}{1-x}$

We have,
$\underset{n\to \infty }{lim}\left\{\right(1+x\left)\right(1+x^2\left)\right(1+x^4)....(1{+}^{2n}\left)\right\}$
$=\underset{n\to \infty }{lim}\left\{\frac{(1-x)(1+x)(1+x^2)....(1+x^{2n})}{1-x}\right\}$
$=\underset{n\to \infty }{lim}\frac{1-x^{4n}}{1-x}$
$=\frac{1-0}{1-x}=\frac{1}{1-x}$ $[\because {lim}_{n\to \infty }{x}^{4n}=0for-1<x<1]$
Hence, option 'B' is correct.